1. ## Help-using exponential distribution

If the number of minutes that a doctor spends with a patient is a random variable having an exponential distribution with the parameter Θ=9, what are the probabilities that it will take the doctor at least 20 minutes to treat
a) One patient;
b) Two patients;
c) Three patients?
Please explain it step by step, thanks!!!

2. Originally Posted by Yan
If the number of minutes that a doctor spends with a patient is a random variable having an exponential distribution with the parameter Θ=9, what are the probabilities that it will take the doctor at least 20 minutes to treat
a) One patient;
b) Two patients;
c) Three patients?
Please explain it step by step, thanks!!!
a) Calculate Pr(Y1 > 20) using the given distribution.

b) Get the pdf for Y = Y1 + Y2 where Y1 and Y2 are i.i.d. random variables and each has an exponential distribution with the parameter Θ=9. Now calculate Pr(Y > 20).

c) Get the pdf for Y = Y1 + Y2 + Y3 where Y1, Y2 and Y3 are i.i.d. random variables and each has an exponential distribution with the parameter Θ=9. Now calculate Pr(Y > 20).

3. Well, first of all in parts 2 and 3 we need the time to serve the patients to be independent.
Then by the mgf we can obtain the sum of two and three exponentials.
Since an $\displaystyle Exp(\theta)=\Gamma(1,\theta)$
we have the sum of two independent exponentials distributed as a $\displaystyle \Gamma(2,\theta)$
and the sum sum of three independent exponentials distributed as a $\displaystyle \Gamma(3,\theta)$.
Now integrate the corresonding densities.

4. Originally Posted by matheagle
Well, first of all in parts 2 and 3 we need the time to serve the patients to be independent.
Then by the mgf we can obtain the sum of two and three exponentials.
Since an $\displaystyle Exp(\theta)=\Gamma(1,\theta)$
we have the sum of two independent exponentials distributed as a $\displaystyle \Gamma(2,\theta)$
and the sum sum of three independent exponentials distributed as a $\displaystyle \Gamma(3,\theta)$.
Now integrate the corresonding densities.
Can you explain more, or can you write out? I'm still confuse about the part 2 and 3

I'm sure they prove that...
$\displaystyle \Gamma(a,b)+\Gamma(c,b)=\Gamma(a+c,b)$
as long as the two are independent.
The proof is trivial using MGFs. That's the point of this question.
You need this to add two independent Chi-Squares to get another Chi-Square...

6. Originally Posted by matheagle
I'm sure they prove that...
$\displaystyle \Gamma(a,b)+\Gamma(c,b)=\Gamma(a+c,b)$
as long as the two are independent.
The proof is trivial using MGFs. That's the point of this question.
You need this to add two independent Chi-Squares to get another Chi-Square...
that's what i'm really confuse about, can you wrote down the steps so i can figure out how to do it. thanks

7. $\displaystyle \Gamma(a,b)+\Gamma(c,b)=\Gamma(a+c,b)$
assuming independence of course.
Let $\displaystyle X\sim\Gamma(a,b)$ and $\displaystyle Y\sim\Gamma(c,b)$.
Then $\displaystyle E(e^{(X+Y)t})=E(e^{Xt})E(e^{Yt})$.
plugging in our constants...
$\displaystyle E(e^{(X+Y)t})=(1-bt)^{-a}(1-bt)^{-c}=(1-bt)^{-(a+c)}$.
AND that's a $\displaystyle \Gamma(a+c,b)$.

8. Originally Posted by matheagle
$\displaystyle \Gamma(a,b)+\Gamma(c,b)=\Gamma(a+c,b)$
assuming independence of course.
Let $\displaystyle X\sim\Gamma(a,b)$ and $\displaystyle Y\sim\Gamma(c,b)$.
Then $\displaystyle E(e^{(X+Y)t}=E(e^{Xt})E(e^{Yt})$.
So, In part 2, the answer is P(Y>20)= 1-P(Y<20), where the equation changes to ((1/Θ)e^(-x/Θ))^2 with Θ=9?

9. Originally Posted by Yan
So, In part 2, the answer is P(Y>20)= 1-P(Y<20), where the equation changes to ((1/Θ)e^(-x/Θ))^2 with Θ=9?

NO, you need to use the gamma density
And it doesn't matter which region you integrate.
LOOK at the gamma and read what I wrote

10. Originally Posted by matheagle
NO, you need to use the gamma density
And it doesn't matter which region you integrate.
LOOK at the gamma and read what I wrote
but the question is about finding the probabilities, how could you find out the probabilities without the region. Can you just wrote out the answer step by step? so i can follow you step to find out what happen to this problem? Thanks!!

11. Originally Posted by Yan
but the question is about finding the probabilities, how could you find out the probabilities without the region. Can you just wrote out the answer step by step? so i can follow you step to find out what happen to this problem? Thanks!!

I DID.
I put in every step.
Integrate a gamma(2,9) from 20 to infinity.
and a square of a density is not a density
YOU need to review this chapter on changing variables.

12. Originally Posted by matheagle
I DID.
I put in every step.
Integrate a gamma(2,9) from 20 to infinity.
and a square of a density is not a density
YOU need to review this chapter on changing variables.
it is impossible for part 3. the probability with 3 patient is <0.
Can you show all your work? Don't repeat what you said before. thanks

13. Originally Posted by Yan
If the number of minutes that a doctor spends with a patient is a random variable having an exponential distribution with the parameter Θ=9, what are the probabilities that it will take the doctor at least 20 minutes to treat
a) One patient;
b) Two patients;
c) Three patients?
Please explain it step by step, thanks!!!
Do you know the property that relates exponential random variables (or Poisson processes) to Poisson distribution? If you do, then there is an easy way...

Let's denote by $\displaystyle N$ the number of patients that have been taken care of (and have left) during 20 minutes. The fundamental fact is that $\displaystyle N$ has a Poisson distribution of parameter $\displaystyle \lambda=\frac{9}{60}\cdot 20=3$ (Important note: I'm assuming $\displaystyle \Theta$ is 9 "per hour", because an average 9 patients by minute seems huge...).

Therefore,
a) the probability that the first patient is still with the doctor after 20 minutes is $\displaystyle P(N=0)=e^{-\lambda}=e^{-3}\simeq 5\%$,
b) $\displaystyle P(N=1)=\lambda e^{-\lambda}\simeq 15\%$,
c) $\displaystyle P(N=2)=\frac{\lambda^2}{2}e^{-\lambda}\simeq 22\%$.

Isn't that somewhat simpler ? (provided you know the fact that $\displaystyle N$ is a Poisson random variable)